killjoy1990xb9

2022-01-20

Solving simple differential equation
$\left(t+4\right)dx=4\left(1+{x}^{2}\right)dt$

Bob Huerta

As you rearranged it in your post, we have
$\int \frac{dx}{1+{x}^{2}}=4\int \frac{dt}{t+4}$
As you discovered, the LHS is arctan. To find the value of the RHS, we may say . Thus the RHS can be rewritten as
$4\int \frac{du}{u}=4\mathrm{log}|u|+{C}_{2}$
Because we want this integral to be in respect to t, we have $u=t+4$, so
$\mathrm{arctan}x+{C}_{1}=4\mathrm{log}|t+4|+{C}_{2}$
To confirm that the integral we solved is indeed correct, you can simply differentiate $4\mathrm{log}|t+4|+{C}_{2}$
Thus $x=\mathrm{tan}\left(4\mathrm{log}|t+4|+{C}_{2}-{C}_{1}\right)$
and $t=\mathrm{exp}\left(\frac{1}{4}\left(\mathrm{arctan}x+{C}_{1}-{C}_{2}\right)\right)-4$

Dawn Neal

Here are some proceeding steps:
- Seperating the variables you get
$\frac{dx}{1+{x}^{2}}=4\cdot \frac{dt}{t+4}$
- Integrating both sides you have:
${\mathrm{tan}}^{-1}\left(x\right)+{C}_{1}=4\cdot \mathrm{log}|t+4|+{C}_{2}$.

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